Question

two number are selected at random from a set of first 90 natural numbers. find the probability that the products of randomly selected number is divisible by 3 ​

Answer 1

149/267 answer

Steps :

-------1st separate the 90 numbers in two groups

a) multiple of 3 { total no. =30}

b) not multiple of 3 { total no. =60 }

------ I think there be two possibilities

*Either* , select both form group (a)

*Or* , select one from (a) and rest from (b)

------ Finally , I got

Answer 2

Let us assuming the numbers are distinct in nature , which is the  way of reading the question IMO.

Our objective is  would tell us that it must be exactly 50%, that may be erineous.

Let  we picked  the first number already. There are two alternatives :The first number is even (50%).

Then, there are 4 even numbers that are favoured  out of 9 left. So this is estimate at  50%×44%=22%.The first number is odd (50%).

Then, there are 4 odd numbers that are considered favourable out of 9 left.

So this is estimated at  50%×44%=22%. total, the probability is 44% (49 to be exact).

Alternative solution:There are (102)=45 possible waysis  to choose 2 numbers.

Choosing two odd numbers can take place in (52)=10 ways, and the same is true for picking  two even numbers.

So in total, we have 20 favourable outcomes.Therefore , the final result is p=2⋅(52)(102)=2045=49=44%  ……………………